OFFSET
0,3
FORMULA
Given g.f. A(x), let G(x) be defined by G(x*A(x)^2) = x, then
(1) G(x) = x/(1 + A(x)^2*G(x))^2,
(2) A(G(x)) = 1 + A(x)^2*G(x).
From Seiichi Manyama, Mar 01 2025: (Start)
Let a(n,k) = [x^n] A(x)^k.
a(n,0) = 0^n; a(n,k) = k * Sum_{j=0..n} binomial(2*n-2*j+k,j)/(2*n-2*j+k) * a(n-j,2*j). (End)
EXAMPLE
G.f. A(x) = 1 + x + 2*x^2 + 9*x^3 + 52*x^4 + 372*x^5 + 3058*x^6 +...
A(x)^2 = 1 + 2*x + 5*x^2 + 22*x^3 + 126*x^4 + 884*x^5 + 7149*x^6 +...
A(x*A(x)^2) = 1 + x + 4*x^2 + 22*x^3 + 156*x^4 + 1285*x^5 + 11886*x^6 +...
A(x*A(x)^2)^2 = 1 + 2*x + 9*x^2 + 52*x^3 + 372*x^4 + 3058*x^5 +...
Define G(x) by G(x*A(x)^2) = x, then
G(x) = x - 2*x^2 + 3*x^3 - 12*x^4 + 17*x^5 - 198*x^6 - 345*x^7 +...
such that G(x) = x/(1 + A(x)^2*G(x))^2.
PROG
(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=0, n, A=1+x*subst(A^2, x, x*A^2)); polcoeff(A, n)}
(PARI) a(n, k=1) = if(k==0, 0^n, k*sum(j=0, n, binomial(2*n-2*j+k, j)/(2*n-2*j+k)*a(n-j, 2*j))); \\ Seiichi Manyama, Mar 01 2025
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Paul D. Hanna, Aug 21 2008
STATUS
approved